/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be disproven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) NonTerminationLoopProof [COMPLETE, 40 ms]
(4) NO


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(0, 1, X) -> h(X, X)
   h(0, X) -> f(0, X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, 1, X) -> H(X, X)
   H(0, X) -> F(0, X, X)

The TRS R consists of the following rules:

   f(0, 1, X) -> h(X, X)
   h(0, X) -> f(0, X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = H(g(X, 0), g(1, Y)) evaluates to  t =H(g(1, Y), g(1, Y))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [X / 1, Y / 0]

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Rewriting sequence

H(g(1, 0), g(1, 0)) -> H(0, g(1, 0))
with rule g(X, Y') -> Y' at position [0] and matcher [X / 1, Y' / 0]

H(0, g(1, 0)) -> F(0, g(1, 0), g(1, 0))
with rule H(0, X') -> F(0, X', X') at position [] and matcher [X' / g(1, 0)]

F(0, g(1, 0), g(1, 0)) -> F(0, 1, g(1, 0))
with rule g(X', Y) -> X' at position [1] and matcher [X' / 1, Y / 0]

F(0, 1, g(1, 0)) -> H(g(1, 0), g(1, 0))
with rule F(0, 1, X) -> H(X, X)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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(4)
NO